2sin^2x - 5sinx + 3 < 0

[tdotgirl]

2sin^2x - 5sinx + 3 < 0

this is how i am trying to solve it.


1. first i factored it:
(2sinx - 1) (sinx - 3) < 0


2. then i set (2sinx - 1) and (sinx - 3) equal to zero:

2sinx - 1 = 0
sinx = 1/2
x = pi/6, 5pi/6

sinx - 3 = 0
sinx = 3 (no solution)


3.


<-------------|----------|-----------------------|-------->
0 pi/6 5pi/6

now i find which x's give a negative value in the intervals (-infinity, pi/6), (pi/6, 5pi/6), and (5pi/6, infinity)



is this how i'm supposed to do it?

 

[skeeter]

if you want to do the entire set of Reals, then sinx = 1/2 has an infinite number of solutions, not just pi/6 and 5pi/6.

restrict your solutions to values of x between 0 and 2pi.